A 0.10M solution of a certain monoprotic weak acid has a pH of 2.44. Calculate Ka.
.............HA ==> H^+ + A^-
I...........0.1.....0.....0
C...........-x......x.....x
E.........0.1-x.....x.....x
The problem tells you pH = 2.44. Convert that to (H^+) by pH = -log(H^+) and substitute the x value in for (H^+) and (A^-). Evaluate 0.1-x and substitute that for (HA) in the Ka expression and evaluate Ka.
To calculate the Ka (acid dissociation constant) of a weak acid, we need to use the Henderson-Hasselbalch equation. The Henderson-Hasselbalch equation relates the pH, pKa, and the ratio of the concentration of the conjugate base to the concentration of the weak acid.
The equation is given as:
pH = pKa + log([A-]/[HA])
Where:
pH = the known pH of the solution
pKa = the negative logarithm of the Ka
[A-] = concentration of the conjugate base
[HA] = concentration of the weak acid
In this case, we have a 0.10 M solution of the weak acid with a pH of 2.44. We can rearrange the Henderson-Hasselbalch equation to solve for pKa:
pKa = pH - log([A-]/[HA])
Now, let's substitute the given values into the equation:
pKa = 2.44 - log([A-]/[HA])
Since the weak acid is monoprotic, we can assume that the concentration of the conjugate base ([A-]) is equal to the concentration of the weak acid ([HA]).
pKa = 2.44 - log(1)
log(1) is equal to 0, so the equation simplifies to:
pKa = 2.44
Finally, to find the Ka, we take the antilog of the negative pKa:
Ka = 10^(-pKa)
Ka = 10^(-2.44)
Using a calculator, we find:
Ka ≈ 3.63 x 10^(-3)
Therefore, the Ka value for the weak acid is approximately 3.63 x 10^(-3).